🧰 Mastering NEC 392.22(B)(1)(c): Proper Cable Tray Sizing for Mixed Single Conductors

When you’re installing single-conductor cables in a ladder-type cable tray, and you’re mixing large conductors (≥1000 kcmil) with smaller ones (<1000 kcmil), the National Electrical Code (NEC) requires a special fill calculation method under Article 392.22(B)(1)(c).

This guide walks you through exactly what the code means, how to apply it, and how to calculate the correct tray size using simple steps and accurate math.

⚖️ What NEC 392.22(B)(1)(c) Says

“Where 1000 kcmil or larger single-conductor cables are installed in the same cable tray with single-conductor cables smaller than 1000 kcmil, the sum of the cross-sectional areas of all cables smaller than 1000 kcmil shall not exceed the maximum allowable fill area resulting from the computation in Column 2 of Table 392.22(B)(1) for the appropriate cable tray width.”

Key Takeaways:

• This rule only applies when both ≥1000 kcmil and <1000 kcmil single-conductor cables are in the same tray.

• Only the smaller conductors (<1000 kcmil) are subject to the fill limit.

• The larger conductors do not count in fill, but they reduce how much room is left for smaller ones.

🧮 The Formula: Column 2 Adjustment

When this rule applies, you must use an adjusted fill calculation from Column 2 of Table 392.22(B)(1). The NEC provides this formula:

Max Fill = Base − (1.1 × Sd)

Where:

• Base = Column 2 value based on tray width (see table below)

• Sd = Space demand (in inches) occupied by the larger conductors (number of ≥1000 kcmil conductors × their outside diameter)

• 1.1 = Fixed reduction factor that applies across all tray sizes

This adjustment reflects the physical space the larger conductors take up, reducing how much room is left for smaller conductors in the tray.

📊 Column 2 – Simplified Fill Table

👉 Column 2 is used ONLY for small conductors (<1000 kcmil) when mixed with ≥1000 kcmil conductors. You do not use Column 2 for normal tray sizing.

🧾 Example: (5) 1/0 THHN + (5) 1000 kcmil THHN in the Same Tray

Let’s size a tray for the following:

• (5) × 1/0 AWG THHN (red)

• (5) × 1000 kcmil THHN (black)

Step 1: Determine Area of 1/0 THHN

From NEC Chapter 9, Table 5:

• 1/0 THHN OD = 0.545″

• Area = 0.1855 in²

Total area used = 5 × 0.1855 = 0.9275 in²

Step 2: Determine Space Demand (Sd) of 1000 kcmil

Assume:

• 1000 kcmil THHN OD = 1.2″

• Sd = 5 × 1.2 = 6.0 inches

📏 Step 3: Check Various Tray Widths Using the Formula

🔸 6-Inch-Wide Tray

Base = 6.5 in²
Max Fill = 6.5 − (1.1 × 6.0) = 6.5 − 6.6 = −0.1 in²
Fill required = 0.9275 in²
❌ Not Compliant

🔸 12-Inch-Wide Tray

Base = 13.0 in²
Max Fill = 13.0 − (1.1 × 6.0) = 13.0 − 6.6 = 6.4 in²
Fill required = 0.9275 in²
✅ Compliant

🔸 18-Inch-Wide Tray

Base = 19.5 in²
Max Fill = 19.5 − (1.1 × 6.0) = 19.5 − 6.6 = 12.9 in²
Fill required = 0.9275 in²
✅ Compliant with extra margin

🛠️ Key Takeaways

• Use Column 2 only when small conductors (<1000 kcmil) are in the same tray as larger conductors (≥1000 kcmil).

• Larger conductors do not count toward fill, but reduce the fill allowance for the smaller ones.

• The formula Max Fill = Base − (1.1 × Sd) ensures you don’t overcrowd the tray.

• Use correct cross-sectional areas from NEC Chapter 9, Table 5 (e.g., 1/0 THHN = 0.1855 in²).

⚠️ Pro Tips

• Always verify outside diameter from manufacturer specs or NEC Table 5.

• When practical, use separate trays for large and small conductors to avoid this complexity.

• Check with your AHJ to confirm local enforcement of 392.22(B)(1)(c).

✅ Final Thoughts

NEC 392.22(B)(1)(c) is a special rule that kicks in only when mixing large and small single conductors in the same tray. You don’t count the big ones in the fill — but you do reduce the fill allowance for the smaller ones using a clear formula.

By using the correct conductor dimensions and following the simplified Column 2 logic, you can ensure your tray installations are safe, efficient, and fully code-compliant.